Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $g \in L^{\infty}(\mathbb R)$. Consider the operator $$ \begin{split} T_g\colon & L^2(\mathbb R)\to L^2(\mathbb R) \\ & f \mapsto gf \end{split} $$ Prove that $T_g$ is compact (i.e., the image under $T_g$ of bounded closed sets is compact) if and only if $g=0$ a.e.

I do not know how to start and I'm very puzzled. I know very little about compactness in $L^p$: of course they are complete metric spaces, therefore a subspace is compact if and only if it is closed (complete) and totally bounded. A singleton is of course totally bounded and I think it is closed: therefore I can say that if $g=0$ a.e. then the image of every subspace is $\{0\}$ which is compact, so the operator is compact. What about the inverse direction? It seems hard to prove. Would you help me, please? Thanks.

share|improve this question

3 Answers 3

up vote 7 down vote accepted

Suppose that $g$ is not zero a.e. and let $\epsilon>0$ be such that $E=\{x:|g(x)|>\epsilon\}$ has nonzero measure. Consider the orthogonal projection $p=T_{\chi_{E}}$. Assume that $T_{g}$ is a compact operator. The operator $T_{g}$ naturally induces a continuous linear operator of $L^2(\mathbb{R})$ into the Hilbert subspace $pL^{2}(\mathbb{R})=\{\xi \in L^{2}(\mathbb{R}):p\xi=\xi\}$, which is a Banach space. This map is onto, since $\xi \in pL^{2}(\mathbb{R})$ is mapped to by the $L^{2}$ function equal to $\xi(x)/g(x)$ on $E$ and zero elsewhere. By the open mapping theorem we have that this induced map is open, and therefore the image of the unit ball of $L^{2}(\mathbb{R})$ under this induced map contains an open ball $B$, and therefore the closure contains a closed ball $\overline{B}$. This closed ball is not compact, since $pL^2(\mathbb{R})$ is infinite-dimensional. But the image of the unit ball of $L^2(\mathbb{R})$ under a compact operator must have compact closure, and closed subsets of compact sets must be compact, so this is a contradiction.

share|improve this answer
    
Thank you for your reply. One question: what is $B(L^2(\mathbb R))$? The unit ball in $L^2$, isn't it? –  Romeo Jul 25 '12 at 13:47
    
Hi Romeo: B(L^2(R)) is the set of all bounded operators from L^2(R) to L^2(R). –  Jon Bannon Jul 25 '12 at 14:38
    
Thanks. Sorry, but I don't understand how you define the projection $p$: what does $p$ send a bounded operator $L^2\to L^2$ to? –  Romeo Jul 25 '12 at 16:13
    
Look at the characteristic function of the support of g. The multiplication operator associated to the characteristic function (just like the one you defined) is the projection p. –  Jon Bannon Jul 25 '12 at 16:17
1  
@JonBannon: No I don't think there is a problem with measure classes. You say: "The operator $T_g$ naturally induces a continuous linear operator of $L^2(\mathbb R)$ onto the Hilbert subspace $pL^2(\mathbb R)$". I just don't see why it need be onto. Probably you want to pick $\varepsilon$ such that $p = 1_{ \{ x \in \mathbb R \mid |g(x)| > \varepsilon \} }$ is non-zero. Then the induced map will be onto. –  Jesse Peterson Aug 15 '12 at 1:58

Suppose, there is some $\epsilon > 0$ such that $M_\epsilon = \{ x \;\vert\; g(x) > \epsilon\}$ has positive measure $\mu(M_\epsilon) > 0$. Now pick a sequence of sets $M_n \subset M_\epsilon$ with

  1. $M_{n+1} \subset M_n$ and
  2. $\mu(M_{n+1}) < \frac{1}{2}\mu(M_n)$ for all $n \in \mathbb N$.

Note that 2. implies $\mu(M_n) > 0$ for all $n\in \mathbb N$ and $\mu(M_n) \to 0$ as $n \to \infty$. Let $f_n(x) = \mu(M_n)^{-1}\chi_{M_n}(x)$ denote the normalized characteristic function of $M_n$. We will show now, that $T_gf_n$ does not contain a convergent subsequence:

We have

$$\Vert T_gf_m - T_gf_n\Vert_2^2 \geq \epsilon^2\Vert f_m -f_n \Vert_2^2 = \epsilon^2 \left(\mu(M_m)\left( \frac{1}{\mu(M_m)} - \frac{1}{\mu(M_n)}\right)^2 + \frac{\mu(M_n \setminus M_m)}{\mu(M_n)^2} \right)\geq \epsilon^2 \frac{\mu(M_n)}{\mu(M_n)^2} \to \infty$$

for $m > n$ and $\mu(M_n) < 1$.

share|improve this answer

If $\lVert g\rVert_{\infty}\neq 0$, fix, $\Delta_n$ a sequence of measurable subsets of $\Bbb R$ such that $\lambda(\Delta_n)\in (0,+\infty)$ and for all $n\in \Bbb N$, $x\in\Delta_n$, we have $|g(x)|\geq\lVert g\rVert_{\infty}-\frac 1n$.

  • We can assume that $\lambda(\Delta_n)\to 0$, since the sequence $\{\Delta_n\}$ can wbe chosen decreasing, and if the measure of the intersection is non-negative, $g$ would be constant equal to $C$ over a set of positive measure $A_0$. If we take a sequence of the form $\chi_{A_0}f_n$ which is weakly convergent but non fo the norm. Then $T_g(\chi_{A_0}f_n)=C\chi_{A_0}f_n$, which is not strongly convergent. Hence in this case we already have that $T_g$ is not compact.
  • We would have, if $T_g$ were compact, that
    $$\tag{*} \lim_{N\to +\infty}\sup_{n\in\Bbb N}\int_{\{1\geq N\lambda(\Delta_n)\}}g^2\frac{\chi_{\Delta_n}}{\lambda(\Delta_n)}dx=0.$$ To see this, use precompactness and a "$2\varepsilon$" argument. With the assumption made in the first point, we can, for each integer $N$, pick an integer $k(N)$ with $1\geq N\lambda(\Delta_{k(N)})$. We can choose the sequence $\{k(N)\}$ strictly increasing, hence $1/k(N)< \lVert g\rVert_{\infty}$ for $N$ large enough. We would have $$\lim_{N\to +\infty}\left(\lVert g\rVert_{\infty}-\frac 1{k(N)}\right)^2.$$ But this limit is $\lVert g\rVert_{\infty}$.

An alternative method is the following: we use the fact that the spectrum of a multiplication operator is the essential range of $g$. The spectrum of a compact self-adjoint operator is a sequence which converge to $0$, hence the essential image of $g$ is $\{0\}$.

share|improve this answer
    
Thanks for your help, Davide. May I ask you some more details about the "alternative method"? Why the spectrum of a multiplication operator is the essential range of $g$? Do you have any references? (As you have surely understood, I haven't studied functional analysis yet, I'm studying it by myself). Thank you very much indeed. –  Romeo Jul 25 '12 at 19:18

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.